Monthly payment calculator not working out

[FuzzyMidget]

Hey again guys,

I'm still learning here so bear with me. I'm trying to start this lightweight payment calculator, but it's giving me a problem when I enter data. After I scanf to principal, term and rate, if I printf to check the new values, the first two are correct. However, it shows the value of rate after the scanf to be 0.00 regardless of what number I enter. I suspect the cause will be obvious, but I'm just not seeing it. I'd appreciate a little help pinpointing it.

#include <stdio.h>
#include <stdlib.h>
#include <ctype.h>


int main()
{
	double moPay, rate, principal, term;
	int months;
/*	- moPay is the monthly paymount amount to be calculated
	- rate is the interest rate
	- principal is the balance due before interest
	- term is the number of years the payments will last
	- months is the the term converted into months by multiplying by 12 
*/
	
	printf("\nEnter the amount of the loan using\ndecimal places (ex: $1000 is 1000.00): ");
	scanf("%lf", &principal);
	printf("\nNow enter the term of the loan in years: ");
	scanf("%d", &term);
	printf("\nEnter the interest rate for the loan (example: 5%% is 5): ");
	scanf("%lf", &rate);


	printf("\nPrincipal is %.2lf, term is %d, and rate is %lf.\n", principal,term,rate);
	
system("pause");
return 0;
}

[Salem]

> printf("\nPrincipal is %.2lf, term is %d, and rate is %lf.\n", principal,term,rate);
Because %d is not what you use to print a floating point value.

[Matticus]

scanf("%d", &term);

Same for reading a floating point value. Did you mean to declare "term" as an int?

[FuzzyMidget]

scanf("%d", &term);

Same for reading a floating point value. Did you mean to declare "term" as an int?

Ahh, that's it. Yes I did intend for term to be an int. I gotta start paying better attention to these things. Thanks a lot guys.

[Matticus]

Indeed you do. But even the best of us have slip-ups. That's why compiler warnings are so very important. For instance, compiling your code, I got:

main.c||In function 'main':|
main.c|20|warning: format '%d' expects type 'int *', but argument 2 has type 'double *'|
main.c|25|warning: format '%d' expects type 'int', but argument 3 has type 'double'|
main.c|9|warning: unused variable 'months'|
main.c|8|warning: unused variable 'moPay'|
||=== Build finished: 0 errors, 4 warnings ===|

The first two warnings explicitly mention the type of mistake and the associated line number.

So you should heed all compiler warnings. If you didn't see these while compiling, then you likely need to increase your compiler warnings (or, if you're using certain IDEs, don't "build and run", but just "build" to see the warnings).

[FuzzyMidget]

@Matticus, thanks for the input, and I actually only got one "unused variable" warning originally. I've gotten a little more into it and am having issues with the payment calculation now. It's coming back with payments that are way too low (not even half what they should be). I'm also getting a different warning now about "call to 'pow' with no prototype in main". Any chance someone can point out where I'm dropping the ball on this?

Here's the newest code:

#include <stdio.h>
#include <stdlib.h>
#include <ctype.h>




int main()
{
    double moPay, annRate, principal, rate, ratePow, totalPmt, totalInt; 
    int months, yrs, term; 
/*    - moPay is the monthly paymount amount to be calculated
    - annRate is the annual interest rate
    - principal is the loan amount before interest
    - yrs is the number of years the loan is for
    - term is the life of the loan converted into months by multiplying yrs by 12 
*/
    /* user input of loan data */
    printf("\nEnter the amount of the loan: ");
    scanf("%lf", &principal);
    printf("\nNow enter the life of the loan in years: ");
    scanf("%d", &yrs);
    printf("\nEnter the annual interest rate for the loan (example: 5%% is 5): ");
    scanf("%lf", &annRate);


    /* loan and payment calculation */
    term=yrs  * 12;
    rate=annRate / 1200.0;
    moPay=(rate+rate / (pow (1.0+rate, term)-1.0)) * principal;    
    
    totalPmt=term * moPay;
    totalInt=totalPmt - principal;
    
    /* output loan and payment details */
    printf("\nPrincipal is %.2lf, term is %d months, and rate is %.0lf%%.\n", principal,term,annRate);
    printf("\nYour monthly payment is %.2lf.\n\nYour total interest is %.2lf.\n\nYour total payment is %.2lf.\n", moPay,totalInt,totalPmt);
    
    
    
system("pause");
return 0;
}

[Hodor]

Where is the prototype for pow()? I suspect that it's in <math.h> (so #include <math.h>)

Something else to consider is how C deals with a function that's called where there is no prototype for that function "scope"; i.e. in your code the C compiler has no idea what pow() returns or what parameters it expects because the prototype is in math.h which you haven't included. The answer is that the compiler makes some assumptions. First, it doesn't make any guess about the number of parameters at all (maybe pow() expects 5000 parameters, or maybe it expects none... the C compiler will not even attempt to guess this, it's can't). The second assumption that the compiler will make is that pow() returns an int. It just so happens that pow() does not return an int, but the compiler makes that assumption anyway just because there really isn't any better option. This has implications. Looking at this line:

moPay=(rate+rate / (pow (1.0+rate, term)-1.0)) * principal;

The compiler has tried to make an educated guess that the pow() function returns an int. But, in reality, pow() returns a double. But because the compiler thinks it returns an int (due to the lack of a function prototype) the return value of pow() loses the decimal parts of the floating point number with obviously bad results.

[FuzzyMidget]

@Hodor, thanks for that. Just including math.h fixed everything. Thanks again.